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Q.
The value of the integral $\int^{b}_{a}\frac{\left|x\right|}{x}=dx,$ where $a < b$ is
Integrals
Solution:
Three cases may arise
Case (i) : let $0 \le a < b,$ then $\frac{\left|x\right|}{x}=\frac{x}{x}=1$
$\therefore \int^{b}_{a} \frac{\left|x\right|}{x}dx=\int^{b}_{a}dx=b-a=\left|b\right|-\left|a\right|$
$\left(Since |a| = a \,and |b| = b\right)$
Case (ii) : Let $a < b \le0$, then $\frac{\left|x\right|}{x}=\frac{-x}{x}=-1$
$\therefore \int^{b}_{a} \frac{\left|x\right|}{x}dx=\int^{b}_{a}dx=\int^{b}_{a}\left(-1\right)dx=a-b=-\left|a\right|+\left|b\right|$
$\left(Since |a| = -a \,and |b| = -b\right)$
Case (iii) : Let a < b < 0, then
$\int^{b}_{a}\frac{\left|x\right|}{x}=dx= \int^{b}_{a} \frac{\left|x\right|}{x}dx=\int^{b}_{a} \frac{\left|x\right|}{x}dx$
$=\int^{b}_{a}\left(-1\right)dx=\int^{b}_{a} \left(1\right)dx=a+b=-\left|a\right|+\left|b\right|
\left(Since |a| = -a \,and |b| = b\right)$
Thus, in every case $\int^{a}_{b} \frac{\left|x\right|}{x}dx=\left|b\right|-\left|a\right|$